Question

$Iff\left(x\right)=\left(lo{g}_{cotx}tanx\right)(lo{g}_{tanx}cotx{)}^{-1}+ta{n}^{-1}\left(\frac{x}{\sqrt{(4-x^2)}}\right)$ then f'(0) is equal to

• A. $-2$
• B. $2$
• C. $\frac{1}{2}$
• D. $0$

Answer 1

The correct option is C $\frac{1}{2}$

$f\left(x\right)=\frac{\left(lo{g}_{cotx}tanx\right)}{\left(lo{g}_{tanx}cotx\right)}+ta{n}^{-1}\left(\frac{x}{\sqrt{(4-x^2)}}\right)$
$f\left(x\right)=1+ta{n}^{-1}\left(\frac{x}{\sqrt{(4-x^2)}}\right)$
Now Put $x=2sin\theta$, we get
$f\left(x\right)=1+ta{n}^{-1}\left(\frac{2sin\theta }{\sqrt{4-4si{n}^2\theta }}\right)\Rightarrow f\left(x\right)=1+ta{n}^{-1}\left(\frac{2sin\theta }{2cos\theta }\right)\Rightarrow f\left(x\right)=1+ta{n}^{-1}\left(tan\theta \right)\Rightarrow f\left(x\right)=1+\theta \Rightarrow f\left(x\right)=1+si{n}^{-1}\left(\frac{x}{2}\right)\Rightarrow f^{\prime }\left(x\right)=0+\frac{1}{\sqrt{1-(\frac{x}{2}{)}^2}}\cdot \frac{1}{2}\Rightarrow f^{\prime }\left(0\right)=\frac{1}{2}$